Harmonic Ratio of the Icosahedron, ~1.2584
From Euclid XIII.16 (Heath, v.3, p. 481)
we have the side as a minor,
[1] S = (R/5)*sqrt(10*(5-sqrt(5))) = m*R/5
where m is the minor, sqrt(10*(5-sqrt(5))),
which is about 5.257311.
Thus, S ~= (1.05)*R: rather close!
Again, to relate S and r (and thus R and r)
we will resort to the central
triangle.
The central triangle argument for the icosahedron
Again, this triangle connects the center
of the icosahedron to a vertex, the side R, to the center
of a equilateral triangular face, the side r,
and across the triangular face, side c.
Applying the Pythagorean theorem to this triangle, we have,
[2] R2 = r2 + c2,
In this case, c is the distance from the center of a equilateral triangle
to one of its vertices. This distance is related to S,
the edge of the triangle, by
[3] S = sqrt(3)*c
as we have seen before, in the discussion of the
equilateral triangle,
and also in the case of the
tetrahedron,
equation [3].
So now, substituting [3] in [1] and squaring, we find,
[4] c2 = S2/3 = (m*R)2*/75
Finally, putting this expression in place of c2 in
equation [2] and collecting terms in R2 on the left,
we have,
[5] {1-m2/75}*R2 = r2
so the harmonic ratio we seek is,
[6] R/r = 1/sqrt(1-m2/75)
which evaluates to 1.258408, rounded to the sixth decimal place as usual.
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